Find where f(x)=4x-tanx,-π/2 < x < π/2 is increasing or decreasing and find its maximum and minimum values.
Answers
function , f(x) = 4x - tanx , -π/2 < x < π/2
differentiating f(x) with respect to x,
f'(x) = 4 - sec²x
now, f'(x) = 4 - sec²x = 0
⇒4 - sec²x = 0
⇒sec²x = (2)²
⇒secx = ±2
⇒x = ±π/3
so, there are three intervals ; (-π/2, -π/3), (-π/3,π/3), (π/3, π/2)
let's check f'(x) > 0 in which intervals.
-π/3 < x < π/3 ⇒secx < 2
so, f'(x) = 4- sec²x > 0
hence, f(x) is increasing in (-π/3, π/3).
similarly, check f'(x) value in (-π/2, -π/3) and (π/3, π/2).
you will get, f'(x) < 0 in these intervals.
so, f(x) is decreasing in (-π/2, -π/3) U (π/3, π/2).
again differentiate with respect to x,
f"(x) = 0 - 2sec²x. tanx = -2sec²x.tanx
at x = π/3, f"(π/3) = -2sec²(π/3).tan(π/3) < 0
hence, f(x) is maximum at x = π/3
and maximum value of f(x) = f(π/3) = 4(π/3) - tan(π/3) = 4π/3 - √3
at x = -π/3, f"(x) = -2sec²(-π/3).tan(-π/3) > 0
hence, f(x) is minimum at x = -π/3
and minimum value of f(x) = f(-π/3) = 4(-π/3) - tan(-π/3) = -4π/3 + √3